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The replacement game (Posted on 2021-03-09) Difficulty: 3 of 5
The triplet (n-1, n, n+1) is written on a board where n is positive integer. A move consists of choosing two of the numbers a and b, replacing them with 2a-b and 2b-a thus creating a new triplet. After a succession of moves, two of the numbers are zeros. Find all possible values for n.

No Solution Yet Submitted by Danish Ahmed Khan    
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Observations, some values | Comment 1 of 2
First some observations:  
A move doesn't change the sum of the triplet = 3n.
Any triplet will maintain the number of even/odd numbers.
The starting position has either one or two odds where the final position (0,0,3n) has either zero or one odds.

This means n is odd.

Also, a number can be factored out of a triplet without affecting whether it will work: (cx,cy,cz) = c(x,y,z)

You can work backwards from any position (x,y,z) to find the three positions that lead to this:  (2x+y, x+2y, 3z)/3 or (2x+z, 3y, x+2z)/3 or (3x, 2y+z, y+2z)/3.

This leads to a lot of branches.  I have found one line of branches that do work, but I don't know if there are others.

My solutions so far consist of n=3^m

(3^m-1, 3^m, 3^m+1) -> (3^m-3, 3^m, 3^m-3) = 3*(3^(m-1)-1,3^(m-1),3^(m-1)+1
continuing to keep the second number unchanged and factoring out the 3 eventually leads to 3^m*(0,1,2) -> 3^m*(0,0,3)

For example without factoring
(26,27,28) ->
(24,27,30) -> 
(18,27,36) ->
(0,27,54) ->
(0,0,81)


Again, I have not shown there are no other solutions, but no other branches look promising.

  Posted by Jer on 2021-03-09 10:15:09
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